# Solutions to the interacting Dirac equation

The Pauli equation can be obtained by taking the non relativistic limit of the interacting Dirac equation (https://en.wikipedia.org/wiki/Pauli_equation, 2). In the process, you remove a phase of $$e^{-imc^{2}/\hbar}.$$ Starting from the solution to the Dirac equation (positive energy and spin Dirac equation solution) $$\Psi_{\uparrow,\mathbf{p}}^{+}(\mathbf{r},t)=\left(\begin{array}{c} 1 \\ 0 \\ \frac{cp_{z}}{\mathcal{E}_{p}+mc^{2}} \\ \frac{c(p_{x}+ip_{y})}{\mathcal{E}_{p}+mc^{2}} \end{array} \right)e^{i(\mathbf{p}\cdot\mathbf{r}-\mathcal{E}_{p}t)/\hbar}.$$ In the non relativistic limit, $$\mathcal{E}=mc^{2}\sqrt{1+\frac{p^{2}}{m^{2}c^{2}}}\approx mc^{2}+\frac{p^{2}}{2m}.$$ Hence the overall phase in the Dirac equation can be expressed as follows: $$e^{-i(\textbf{p}\cdot\textbf{r}-\mathcal{E}t)/\hbar}\approx e^{-imc^2/\hbar} \exp\left(\!-\frac i\hbar \left[\textbf{p}\cdot\textbf{r}+\frac{p^2}{2m}\,t\right]\right).$$ In the derivation of the Pauli equation, you cancel the phase explicitly containing the rest mass energy. Since the only explicit time dependence in the wave function after this cancellation is in the phase, which is now expressed in terms of kinetic energy, applying the energy operator of the Pauli equation gives $$i\hbar\,\partial_{t}\psi=\frac{p^{2}}{2m}\,\psi.$$ For a free particle, the energy operator is giving the total energy. However, clearly if there is potential energy, the energy operator still will only return kinetic energy, and not the potential. So, my question is related to solutions to the interacting Dirac equation; what does $$\mathcal{E}$$ in the phase represent? Should it be contain extra terms, to account for the potential energy?

In a relativistic context, all energy is kinetic. Potential energies alter the dynamics, which then manifests itself as a change in kinetic energy. The reason is simple: energy is a label that comes from one of the Casimir operators of the Lorentz group ($$P_{\mu}P^{\mu}$$), and the Lorentz group knows nothing about potential energies. It knows only about movement in space-time (i.e, kinetic energy)
• So how would one obtain potential energy from the non relativistic limit of the relativistic dispersion relation? ($\mathcal{E}^{2}=m^{2}c^{4}+P^{2}c^{2}$) Commented Jun 6, 2022 at 5:49